Find the Expected Area

Calculus Level 3

A unit circle is constructed centered at ( 0 , 0 ) (0,0) . A point A A is uniformly and randomly chosen on the circumference of the circle with the constraint that point A A is in the first quadrant.

A rectangle is then drawn with one vertex at A A and the vertex opposite A A at the origin ( 0 , 0 ) (0,0) , as shown in the following image:

What is the expected area of the rectangle?

1 3 \frac{1}{3} 1 π \frac{1}{\pi} 2 4 \frac{\sqrt{2}}{4} 1 2 ln ⁡ 2 \frac{1}{2}\ln{2} π 10 \frac{\pi}{10} 1 2 \frac{1}{2} 1 e \frac{1}{e} 1 4 \frac{1}{4}

Nick Turtle by Nick Turtle , 21, USA

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1 solution

Nick Turtle
Dec 10, 2017

To select a point uniformly on the unit quarter circle, it is easiest to select an angle θ \theta between 0 0 and π 2 \frac{\pi}{2} like this:

Then, the height and width of the rectangle is sin ⁡ θ \sin{\theta} and cos ⁡ θ \cos{\theta} , respectively.

Thus, the expected area of the rectangle is 1 π 2 − 0 ∫ 0 π 2 sin ⁡ θ cos ⁡ θ d θ \frac{1}{\frac{\pi}{2}-0}\int_0^{\frac{\pi}{2}}\sin{\theta}\cos{\theta}\ d\theta = 1 π 2 [ 1 2 sin ⁡ 2 θ ] 0 π 2 =\frac{1}{\frac{\pi}{2}}\left[\frac{1}{2}\sin^2{\theta}\right]_0^{\frac{\pi}{2}} = 2 π ⋅ 1 2 =\frac{2}{\pi}\cdot\frac{1}{2} = 1 π =\frac{1}{\pi}

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